Question

root of 1+cosa/1-cosa = coseca +cota

Answer 1

Answer:

this is the answer for your question

Answer 2

Complete question :-

$\sf Prove \: that \: \sqrt{ \dfrac{ 1 + \cos A }{1 - \cos A} } = \mathrm{cosec} A + \cot A$

Solution :-

$\sf \sqrt{ \dfrac{ 1 + \cos A }{1 - \cos A} } = \mathrm{cosec} A + \cot A$

Consider LHS

$\sf \sqrt{ \dfrac{ 1 + \cos A }{1 - \cos A} }$

Muliply both numerator and denominator by √(1 + cos A)

$\sf = \sqrt{ \dfrac{ 1 + \cos A }{1 - \cos A} \times \dfrac{1 + \cos A }{1 + \cos A } } \\\\\\ \sf = \sqrt{ \dfrac{ (1 + \cos A)^2}{1^2- \cos^2 A} } \\\\\\ \sf = \sqrt{ \dfrac{ (1 + \cos A)^2}{1- \cos^2 A} } \\\\\\ \sf = \sqrt{ \dfrac{ (1 + \cos A)^2}{ \sin^2 A} } \\\\\\ \bf \because 1 - \cos^2 A = \sin^2 A \\\\\\ \sf = \dfrac{ \sqrt{(1 + \cos A)^2}}{ \sqrt{\sin^2 A} } \\\\\\ \sf = \dfrac{ 1 + \cos A}{ \sin A} \\\\\\ \sf = \dfrac{1}{ \sin A } + \dfrac{ \cos A}{ \sin A } \\\\\\ \sf = \mathrm{cosec} A + \cot A \\\\\\ \bf \because \dfrac{1}{ \sin A} = \mathrm{cosec} A \: \: and \: \: \dfrac{ \cos A}{ \sin A} = \cot A \\\\\\ \sf = R.H.S$

Hence proved.